Mass of potassium

Question

A patient requires an IV infusion of an isotonic saline solution containing equal molar amounts of potassium chloride and magnesium chloride. In clinical practice, a saline solution is said to be isotonic if it has the same osmotic pressure as blood plasma (7.74 atm at 36 °C). Give the mass of potassium chloride plus the mass of magnesium chloride in grams that needs to be dissolved in 1.42 L of distilled water to make it isotonic. Assume that the change in volume of the solution after adding the salts can be neglected.

Answer 1

Short answer: 6.46 g of KCl plus 8.25 g of MgCl2, which is 14.7 g of salt in total.

The whole problem turns on one idea: osmotic pressure does not count formula units, it counts dissolved particles. Potassium chloride and magnesium chloride do not release the same number of them, so "equal molar amounts" does not mean "equal contributions to the osmotic pressure." Everything below follows from that one sentence.

Step 1 - turn the osmotic pressure into a particle concentration. The van't Hoff equation is pi = i*M*R*T. If you let a single symbol c stand for the total concentration of particles from all sources, it collapses to pi = c*R*T. Temperature must be absolute: 36 deg C = 309.15 K.

c(particles) = pi / (R*T)
             = 7.74 / (0.08206 * 309.15)
             = 7.74 / 25.369
             = 0.3051 mol/L of dissolved particles

That 0.3051 mol/L is the total osmolarity the solution has to hit, and it is worth pausing on, because it is a free sanity check: 0.3051 mol/L is 305 mOsm/L, and real blood plasma runs about 290-300 mOsm/L. Landing right there means Step 1 was done correctly. If your number came out near 30 or near 3000, stop and fix it before doing any chemistry - almost always it is Celsius left in place of Kelvin.

Step 2 - split that concentration between the two salts. Let x be the molarity of each salt, since the problem says equal molar amounts of the two.

1 mol KCl    -> K+  and Cl-             =  2 particles
1 mol MgCl2  -> Mg2+, Cl-, Cl-          =  3 particles
equal moles x of each salt              =  5x particles
5x = 0.3051   ->   x = 0.06102 mol/L

Step 3 - moles, then grams. The instruction to neglect the volume change on dissolving is doing real work here, so do not skip past it: molarity is per liter of solution, not per liter of solvent, and that assumption is exactly what lets you treat the 1.42 L of water as the final solution volume.

n(each salt) = 0.06102 mol/L * 1.42 L
             = 0.08665 mol KCl
             = 0.08665 mol MgCl2

m(KCl)   = 0.08665 mol * 74.55 g/mol =  6.46 g
m(MgCl2) = 0.08665 mol * 95.21 g/mol =  8.25 g
                                       -------
total mass of salt                   = 14.7 g

I used K = 39.10, Mg = 24.31, Cl = 35.45, giving KCl = 74.55 g/mol and MgCl2 = 95.21 g/mol. Use your own table if it differs in the last place; it will not move the third digit of this answer. And with three significant figures in both 7.74 atm and 1.42 L, 14.7 g is the right precision to report.

Or, if you would rather not stop for intermediates, the whole thing is one line:

m(total) = pi * V * (M1 + M2) / (5 * R * T)
         = 7.74 * 1.42 * 169.76 / (5 * 0.08206 * 309.15)
         = 14.7 g

The four errors this problem is built to catch.

forgot the particle count (2 per pair)    ->  36.8 g   (2.5x too much)
used 2 particles for BOTH salts (4/pair)  ->  18.4 g   (1.25x too much)
used T = 36 instead of 309.15 K           ->  about 126 g
read "equal molar" as equal MASSES        ->  wrong split and wrong total

The 18.4 g one is the most worth understanding, because it comes from doing something that feels careful. A student who remembers that salts dissociate, and then applies the same factor of 2 to both of them, has half of the idea: the correction is real, but magnesium chloride releases three ions, not two. The factor is a property of each individual compound, not a blanket rule for "salts."

Two honest footnotes. First, this is the ideal van't Hoff treatment, and real electrolyte solutions fall a little short of it because the ions do not behave completely independently - at this concentration MgCl2 has an osmotic coefficient of roughly 0.9, so a solution actually mixed to this recipe would measure slightly under 7.74 atm. Your course wants the ideal answer, but that gap is the reason a hospital lab measures osmolality rather than calculating it. Second, take the clinical framing lightly: 0.061 M potassium is 61 mmol/L, well above what anyone would infuse, and there is no sodium anywhere in it despite the word "saline." This is a very good osmotic-pressure problem wearing a lab coat.